我们有一个很大的原始数据文件,我们想把它修剪成指定的大小。

如何在python中获取文本文件的前N行?所使用的操作系统对实现有任何影响吗?


当前回答

这里有另一个不错的解决方案与列表理解:

file = open('file.txt', 'r')

lines = [next(file) for x in range(3)]  # first 3 lines will be in this list

file.close()

其他回答

Python 3:

with open("datafile") as myfile:
    head = [next(myfile) for x in range(N)]
print(head)

Python 2:

with open("datafile") as myfile:
    head = [next(myfile) for x in xrange(N)]
print head

下面是另一种方法(Python 2和3都是):

from itertools import islice

with open("datafile") as myfile:
    head = list(islice(myfile, N))
print(head)

没有特定的方法来读取文件对象暴露的行数。

我想最简单的方法是:

lines =[]
with open(file_name) as f:
    lines.extend(f.readline() for i in xrange(N))

如果你想要一些明显(不需要在手册中查找深奥的东西)不需要导入就可以工作的东西,请尝试/except,并且可以在相当大范围的Python 2上工作。X版本(2.2至2.6):

def headn(file_name, n):
    """Like *x head -N command"""
    result = []
    nlines = 0
    assert n >= 1
    for line in open(file_name):
        result.append(line)
        nlines += 1
        if nlines >= n:
            break
    return result

if __name__ == "__main__":
    import sys
    rval = headn(sys.argv[1], int(sys.argv[2]))
    print rval
    print len(rval)

fname = input("Enter file name: ")
num_lines = 0

with open(fname, 'r') as f: #lines count
    for line in f:
        num_lines += 1

num_lines_input = int (input("Enter line numbers: "))

if num_lines_input <= num_lines:
    f = open(fname, "r")
    for x in range(num_lines_input):
        a = f.readline()
        print(a)

else:
    f = open(fname, "r")
    for x in range(num_lines_input):
        a = f.readline()
        print(a)
        print("Don't have", num_lines_input, " lines print as much as you can")


print("Total lines in the text",num_lines)

从Python 2.6开始,您可以利用IO基类中更复杂的函数。所以上面评分最高的答案可以改写为:

    with open("datafile") as myfile:
       head = myfile.readlines(N)
    print head

(你不必担心你的文件少于N行,因为没有StopIteration异常抛出。)